Lecture 2. 2 Graph Isomorphism and Automorphism Groups

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1 Algebra and Computation Lecturer: V. Arvind Lecture 2 Course Instructor: V. Arvind Scribe: Ramprasad Saptharishi 1 Motivation Last lecture we had a brush up of group theory to set up the arsenal required to study Graph Isomorphism. This lecture we shall see how group theory motivates graph isomorphism, and some more theorems on group theory that we would require for later lectures. 2 Graph Isomorphism and Automorphism Groups Recall that two graphs G 1 and G 2 are isomorphic if there is a re-numbering of vertices of one graph to get the other, or in other words, there is an automorphism of one graph that sends it to the other. And clearly, Aut(G) S n, the symmetric group on n objects, which represent the permuation group on the vertices. And since it is a subgroup of the permutation group, Aut(G) n! Of course, providing the entire automorphism group as output would take exponential time but what about a small generating set? Which then leads us to, does there exist a small generating set? Lemma 1. For any group H of size n, there exists a generating set of size log n. Proof. Let H 0 = {e}. If H 0 = H we are done. Otherwise, let x = H \ H 0. Let H 1 = H 0, x and in general, x H \ H i and H = H i, x and since x forms two distinct cosets of H i, H i+1 = 2 H i. And hence, in log n steps one would hit H. Now we can ask the question, can we output a small generating set of the automorphism group of a graph G? We shall refer to this problem as Graph Aut. We shall now show that Graph Iso and Graph Aut are polynomial time equivalent. Theorem 2. With Graph Iso as an oracle, there is a polynomial time algorithm for Graph Aut and vice-versa. 1

2 Proof. First we shall show that we can solve Graph Iso with Graph Aut as an oracle. We are given two graphs G 1 and G 2 and we need to create a graph G using the two such that the generating set of the automorphism should tell us if they are isomorphic or not. Let G = G 1 G 2. Suppose additionally we knew that G 1 and G 2 are connected, then a single oracle query would be sufficient. If any of the generators of Aut(G) interchanged a vertex in G 1 with one in G 2, then connnectivity should force G 1 = G2. But what if they are not connected? We then have this vey neat trick, G 1 = G2 G 2 = G2, and either G 1 or G 1 has to be connected and hence one can check for connectivity and then ask the appropriate query. The other direction is a bit more involved. The idea is to see that any group is a union of cosets. Hence, suppose H = a 1 K a 2 K a n K then {a 1, a 2,..., a n } along with a generating set for K form a generating set for H. Hence once we have a subgroup K with small index, we can then recurse on K. Hence we are looking for a tower of subgroups Aut(G) = H H 1 H 2 H m = {e} such that [H i : H i+1 ] is polynomially bounded. For our graph G, let Aut(G) = H S n. We shall use Weilandt s notation where i π denotes the image of i under π. In this notation, composition becomes simpler: (i π ) τ = i π τ. Define H i = {π H : 1 π = 1, 2 π = 1, i π = 1}. And this gives the tower H 0 = H H 1 H 2 H n 1 = {e} with the additional property that [H i : H i+1 ] n i since there are atmost n i places to go to when the first i are fixed by H i. Now look at the tablaeu Picture supposed to come here, needs to be completed 2

3 3 The Set Stabilizer Problem The Problem Statement: H S n, given by a small generating set. Also given is a subset [n]. Find the stab (H) = {π H : π = } Though this problem has nothing to do with graphs directly, graph isomorphism reduces to this. Theorem 3. Graph Iso P Set Stab Proof. By our earlier theorem, it is enough to show that Graph Aut reduces to Set Stab. One simply needs to note that an automorphism can be thought of as acting on the edges as well. Given a graph G = (V, E), a permutation of the vertices induces a permutation of the edges. Hence, ( ) V φ : Sym(V ) Sym 2 is injective. Thus, all we need to do is find the set of elements in Sym(V ) that stabilizes E. Our set H is φ(sym(v )) Sym ( [( V 2) and = E n 2)] and the automorphism group is precicely stab (H). 4 More Group Theory: Sylow Theorems We will need some more tools for the lectures that follow, the Sylow Theorems in particular. Before that, we need the Orbit-Stabilizer theorem. Definition 4. Let G act on a set S. Let s S The orbit of s (s G ), is the set of all possible images of s under the action of G. s G = {t S : g G, gs = t} The stabilizer of s(g s ) is the set of all elements of G that fix s G s = {g G : gs = s} 3

4 Theorem 5 (Orbit-Stabilizer Theorem). For any finite group G that acts on a set S. For every s S, G = s G G s Proof. This is just Lagrange s theorem, all we need to see is that the stabilizer G s is a subgroup of G and that [G : G s ] = s G, should be a trivial exercise to the reader. Theorem 6 (Sylow Theorems). Let G be a group, G = p m r, where p is a prime and gcd(r, p) = 1. Then 1. there exists a subgroup P such that P = p m (p-sylow subgroup) 2. for any p-subgroup 1 H of G, one of its conjucates is contained in P 3. the number of p-sylow subgroups of G is 1 (mod p) Proof. We shall prove the subdivisions one after another. Subdivision 1: Let Ω be the set of subsets of G of size p m. Note that Ω = ( p m r p m ). Lucas theorem tells us that ( p m r p m ) is not divisible by p by our choice of r. Let G act on Ω by left multiplication. The action decomposes Ω into orbits, and since p Ω, there exists A Ω such that p A G. And since p m r G = A G G A, and by the choice of A, p m G A. And since the elements ga A for a A are distinct under the action of G A, it follows that A G A and hence forcing G A = p m. Hence G A is our desired p-sylow subgroup. Subdivision 2: Let Ω be the set of left cosets of P, our p-sylow subgroup. And let H be a p-subgroup of G, which induces an action on Ω by left multiplication. Since H = p a, every non-trivial orbit of Ω is has cardinality a multiple of p. Hence, the number of points of Ω that are fixed by H is modulo p is the same as Ω. Hence in particular, since p Ω, the set of points fixed by H is non-zero. Hence, there exists a gp that is fixed by H. Hence hgp gp or g 1 hgp P = g 1 hg P for all h H. Thus g 1 Hg P 1 subgroup of order p a for some a 4

5 Note that this also tells us that all p-sylow subgroups are conjucates of each other. Subdivision 3: Let P be a p-sylow subgroup and let Ω be the set of p-sylow subgroups of G on which P acts by conjucation. For any Q Ω, the stabilizer of Q under conjucation is called the normalizer of Q, denoted by N G (Q). Suppose Q Ω is fixed by P on conjucation, then P N G (Q). But subdivision 2 tells us that P and Q are conjucate to each other in N G (Q), which then forces P = Q. Hence the only fixed point is P itself and hence Ω = 1 (mod p). We also did the proof of Lucas Theorem, has to be TEX-ed out 5